Parabolas in standard form!

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Hello, I am having the worst time trying to solve these parabolas and putting them into x-h = a(y-k)^2 form. :(

There are two problems that i keep doing something wrong. could someone solve them so I can have a set up for the rest of my problems? :)

3y^2+6y+108x-969=0

and

x^2+14x=44y+313=0

All help is greatly appreciated!

  • Parabolas in standard form! -

    108x = -3y^2 - 6y + 969
    108x = -3(y^2 + 2y + 1) + 972
    x = -1/36 (y+1)^2 + 9

    I'm not too sure how to do the second one. I'm really sorry.

  • Parabolas in standard form! -

    Can you solve this one?:

    x^2-12x-48y-372=0

    & I made a mistake one the second one!!
    Its actually:

    x^2+14x+44y+313=0 !!

  • Parabolas in standard form! -

    x^2+14x+44y+313=0
    44y = -x^2 - 14x - 313
    44y = -(x^2 + 14x +49-49 - 313
    44y = -( (x+7)^2 - 49 ) - 313
    44y = -(x+7)^2 + 49 - 313
    44y = -(x+7)^2 - 264
    44y + 264 = -(x+7)^2
    44(y + 6) = -(x+7)^2
    y + 6 = (-1/44) (x+7)^2 , in the form that was requested.

    x^2-12x-48y-372=0
    x^2 - 12x - 372 = 48y
    x^2 - 12x +36 = 48y + 372 +36
    (x-6)^2 = 48y + 408
    (x-6)^2 = 48(y + 8.5)

    y + 8.8 = (1/48) (x-6)^2

  • Parabolas in standard form! -

    WOW!

    Thank you so much! you made more sense than my teacher!

    :D

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